Q. 79 - Precalculus Enhanced with Graphing Utilities

Question

In Problems 75– 82, find the difference quotient of $f$; that is, find $\frac{f(x+h)- f\left(x\right)}{h}, h\ne 0,$for each function. Be sure to simplify.

$f\left(x\right)=\frac{1}{x^2}$

Step-by-Step Solution

Answer

The value of $\frac{f(x+h)- f\left(x\right)}{h}=\frac{-2x-h}{x^2\left({x^2+2xh+h}^2\right)}$

Step 1. Given Information

In this problems we have to find the difference quotient of $f$; that is, find $\frac{f(x+h)- f\left(x\right)}{h}, h\ne 0,$for given function. Be sure to simplify.

The given function is
$f\left(x\right)=\frac{1}{x^2}$

Step 2. Firstly we find the value of f ( x + h ) We find the value of f ( x + h ) by putting x = x + h in given function.

$f(x+h)=\frac{1}{{(x+h)}^2}$

Using the formula ${(a+b)}^2=a^2+2ab+b^2$

$f(x+h)=\frac{1}{{x^2+2xh+h}^2}$

 

Step 3. Putting the value of f ( x )   and   f ( x + h ) in the f ( x + h ) -   f ( x ) h

$$\begin{gathered} \frac{f(x+h)- f\left(x\right)}{h}=\frac{\frac{1}{{x^2+2xh+h}^2}-\frac{1}{x^2}}{h} \\ \frac{f(x+h)- f\left(x\right)}{h}=\frac{1}{h}\left(\frac{1}{{x^2+2xh+h}^2}-\frac{1}{x^2}\right) \end{gathered}$$

The LCD of ${x^2+2xh+h}^2 \mathrm{and} x^2 \mathrm{is} x^2\left({x^2+2xh+h}^2\right)$

$$\begin{gathered} \frac{f(x+h)- f\left(x\right)}{h}=\frac{1}{h}\left(\frac{1}{{x^2+2xh+h}^2}·\frac{x^2\left({x^2+2xh+h}^2\right)}{x^2\left({x^2+2xh+h}^2\right)}-\frac{1}{x^2}·\frac{x^2\left({x^2+2xh+h}^2\right)}{x^2\left({x^2+2xh+h}^2\right)}\right) \\ \frac{f(x+h)- f\left(x\right)}{h}=\frac{1}{h}\left(\frac{x^2}{x^2\left({x^2+2xh+h}^2\right)}-\frac{{x^2+2xh+h}^2}{x^2\left({x^2+2xh+h}^2\right)}\right) \\ \frac{f(x+h)- f\left(x\right)}{h}=\frac{1}{h}\left(\frac{x^2-\left({x^2+2xh+h}^2\right)}{x^2\left({x^2+2xh+h}^2\right)}\right) \\ \frac{f(x+h)- f\left(x\right)}{h}=\frac{1}{h}\left(\frac{x^2-{x^2-2xh-h}^2}{x^2\left({x^2+2xh+h}^2\right)}\right) \\ \frac{f(x+h)- f\left(x\right)}{h}=\frac{1}{h}\left(\frac{{-2xh-h}^2}{x^2\left({x^2+2xh+h}^2\right)}\right) \\ \frac{f(x+h)- f\left(x\right)}{h}=\frac{1}{h}·\frac{h(-2x-h)}{x^2\left({x^2+2xh+h}^2\right)} \\ \frac{f(x+h)- f\left(x\right)}{h}=\frac{-2x-h}{x^2\left({x^2+2xh+h}^2\right)} \end{gathered}$$